Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\left(\dfrac{28}{42}+\dfrac{12}{42}-\dfrac{3}{42}\right):\left(\dfrac{-28}{28}-\dfrac{12}{28}+\dfrac{3}{28}\right)\)
\(=\dfrac{37}{42}:\dfrac{-37}{28}=\dfrac{-28}{42}=-\dfrac{2}{3}\)
b: \(=\dfrac{2+8+18+32+50}{12+48+108+192+300}=\dfrac{110}{660}=\dfrac{1}{6}\)
Ta có:
\(D=\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\\ =\dfrac{1.2+1.2.4+1.2.3.3+1.2.2.8+1.2.5.5}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\\ =\dfrac{1.2\left(4+3^2+2.8+5^2\right)}{3.4\left(2^2+3^2+2.8+5^2\right)}\\ =\dfrac{1.2}{3.4}=\dfrac{1}{6}\)
Vậy \(D=\dfrac{1}{6}\)
\(=\frac{1.2\left(1+2.2+3.3+4.4+5.5\right)}{3.4\left(1+2.2+3.3+4.4+5.5\right)}\)
\(=\frac{1.2}{3.4}=\frac{1}{6}\)
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.2.2+2.3.4.2+3.3.6.2+4.3.8.2+5.3.10.2}\)=\(\frac{1}{3.2+3.2+3.2+3.2+3.2}\)=\(\frac{1}{30}\)
vậy bt trên =\(\frac{1}{30}\)
tk nhé
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(=\frac{1.2+2.2.2+3.3.2+4.4.2+5.5.2}{3.4+2.3.2.4+3.3.3.4+4.3.4.4+5.3.5.4}\)
\(=\frac{2.\left(1+2.2+3.3+4.4+5.5\right)}{3.4.\left(1+2.2+3.3+4.4+5.5\right)}\)
\(=\frac{1}{6}\)
= \(\frac{1.2+2.4+3.6+4.8+5.10}{3.2\left(1.2+2.4+3.6+4.8+5.10\right)}=\frac{1}{6}\)
nha :*
10.
\(J=\dfrac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\\ =\dfrac{1\cdot2+2\cdot1\cdot2\cdot2+3\cdot1\cdot3\cdot2+4\cdot1\cdot4\cdot2+5\cdot1\cdot5\cdot2}{3\cdot4+2\cdot3\cdot2\cdot4+3\cdot3\cdot3\cdot4+4\cdot3\cdot4\cdot4+5\cdot3\cdot4\cdot4}\\ =\dfrac{\left(1\cdot2\right)\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}{\left(3\cdot4\right)\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}\\ =\dfrac{1\cdot2}{3\cdot4}\\ =\dfrac{1\cdot1}{3\cdot2}\\ =\dfrac{1}{6}\)
11.
\(K=3^0+3^1+3^2+...+3^{100}\\ =1\cdot\left(3^0+3^1+3^2+...+3^{100}\right)\\ =\dfrac{3-1}{2}\cdot\left(3^0+3^1+3^2+...+3^{100}\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^0+3^1+3^2+...+3^{100}\right)}{2}\\ =\dfrac{3^1-3^0+3^2-3^1+3^3-3^2+...+3^{101}-3^{100}}{2}\\ =\dfrac{3^{100}-3^0}{2}=\dfrac{3^{100}-1}{2}\)
12.
\(L=1-5+5^2-5^3+...+5^{98}-5^{99}\\ =1\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)\\ =\dfrac{5+1}{6}\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)\\ =\dfrac{\left(5+1\right)\cdot\left(1-5+5^2-5^3+...+5^{98}-5^{99}\right)}{6}\\ =\dfrac{5+1-5^2-5+5^3+5^2-5^4-5^3+...+5^{99}+5^{98}-5^{100}-5^{99}}{6}\\ =\dfrac{1-5^{100}}{6}\)
Cảm ơn nhiều nhé!